Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100000;
int L[N + 2];
int Pr[N + 2];
int P[N + 2][19];
vector<int> G[N + 2], nodes[N + 2][19];
void dfs(int node, int pre, int dep) {
  Pr[node] = pre;
  L[node] = dep;
  for (int i : G[node]) {
    if (i == pre) continue;
    dfs(i, node, dep + 1);
  }
}
inline void merge(vector<int>& a, vector<int>& b) {
  for (int i = 0; i < b.size(); i++) a.push_back(b[i]);
  sort(a.begin(), a.end());
  while (a.size() > 10) a.resize(10);
}
void init() {
  for (int i = 1; i <= N; i++) {
    P[i][0] = Pr[i];
    sort(nodes[i][0].begin(), nodes[i][0].end());
    if (nodes[i][0].size() > 10) nodes[i][0].resize(10);
  }
  for (int j = 1; (1 << j) <= N; j++) {
    for (int i = 1; i <= N; i++) {
      P[i][j] = P[P[i][j - 1]][j - 1];
      nodes[i][j] = nodes[i][j - 1];
      merge(nodes[i][j], nodes[P[i][j - 1]][j - 1]);
    }
  }
}
inline int query(int p, int q) {
  if (L[p] < L[q]) swap(p, q);
  for (int i = 18; i >= 0; i--) {
    if (L[P[p][i]] >= L[q]) p = P[p][i];
  }
  if (p < 2) return 1;
  if (p == q) return p;
  for (int i = 18; i >= 0; i--) {
    if (P[p][i] != P[q][i]) {
      p = P[p][i];
      q = P[q][i];
    }
  }
  if (p < 2) return 1;
  return Pr[p];
}
int main() {
  int n, m, q;
  scanf("%d %d %d", &n, &m, &q);
  for (int i = 1; i <= n - 1; i++) {
    int a, b;
    scanf("%d %d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 1; i <= m; i++) {
    int c;
    scanf("%d", &c);
    nodes[c][0].push_back(i);
  }
  dfs(1, 0, 0);
  init();
  while (q--) {
    int u, v, k;
    scanf("%d %d %d", &u, &v, &k);
    if (L[u] > L[v]) swap(u, v);
    int p = query(u, v);
    vector<int> vec;
    for (int i = 18; i >= 0; i--) {
      if (L[v] - (1 << i) >= L[p]) {
        merge(vec, nodes[v][i]);
        v = P[v][i];
      }
      if (L[u] - (1 << i) >= L[p]) {
        merge(vec, nodes[u][i]);
        u = P[u][i];
      }
    }
    merge(vec, nodes[p][0]);
    printf("%d ", min((int)vec.size(), k));
    for (int i = 0; i < k and i < vec.size(); i++) printf("%d ", vec[i]);
    printf("\n");
  }
}